A Little Bit of Complex Analysis

$\begin{aligned} I & = \int_0^1 \log \left(\frac x{x-1}\right) dx & \small \color{#3D99F6} \text{By integration by parts} \\ & = x \log \left(\frac x{x-1}\right) \bigg|_0^1 - \int_0^1 x\left(\frac 1x - \frac 1{x-1}\right) dx \\ & = x \log \left(\frac x{x-1}\right) \bigg|_0^1 + \int_0^1 \frac 1{x-1} dx \\ & = x \log \left(\frac x{x-1}\right) + \log (|x-1|) \bigg|_0^1 \\ & = x \log \left(\frac x{x-1}\right) + \log (1-x) \bigg|_0^1 & \small \color{#3D99F6} \text{See note.} \\ & = \log(-1) & \small \color{#3D99F6} \text{By Euler's formula }e^{i\theta} = \cos \theta + i\sin \theta \\ & = \log \left(e^{i\frac \pi 2}\right) & \small \color{#3D99F6} \text{Considering only the principal branch} \\ & = i\pi \end{aligned}$

$\implies a + b = 1+1 = \boxed{2}$

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Note:

$\begin{aligned} L(1) & = \lim_{x \to 1} \left(x \log x - x \log (x-1) + \log (1-x)\right) \\ & = 0 + \lim_{x \to 1} \log \left(\frac {1-x}{x-1}\right) \\ & = \log (-1) \end{aligned}$

$\begin{aligned} L(0) & = \lim_{x \to 0} \left(x \log x - x \log (x-1) + \log (1-x)\right) \\ & = {\color{#3D99F6} \lim_{x \to 0} \frac {\log x}{\frac 1x}} - 0 + 0 & \small \color{#3D99F6} \text{A }\infty/\infty \text{ case, L'Hôpital's rule applies} \\ & = {\color{#3D99F6} \lim_{x \to 0} \frac {\frac 1x}{-\frac 1{x^2}}} & \small \color{#3D99F6} \text{Differentiate up and down w.r.t. }x \\ & = 0 \end{aligned}$

$\implies L(1) - L(0) = \log (-1)$ .